Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $y \neq 0$. $k = \dfrac{10y - 60}{y^2 - 15y + 54} \times \dfrac{9y^2 - 81y}{y + 2} $
Solution: First factor the quadratic. $k = \dfrac{10y - 60}{(y - 9)(y - 6)} \times \dfrac{9y^2 - 81y}{y + 2} $ Then factor out any other terms. $k = \dfrac{10(y - 6)}{(y - 9)(y - 6)} \times \dfrac{9y(y - 9)}{y + 2} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ 10(y - 6) \times 9y(y - 9) } { (y - 9)(y - 6) \times (y + 2) } $ $k = \dfrac{ 90y(y - 6)(y - 9)}{ (y - 9)(y - 6)(y + 2)} $ Notice that $(y - 6)$ and $(y - 9)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ 90y(y - 6)\cancel{(y - 9)}}{ \cancel{(y - 9)}(y - 6)(y + 2)} $ We are dividing by $y - 9$ , so $y - 9 \neq 0$ Therefore, $y \neq 9$ $k = \dfrac{ 90y\cancel{(y - 6)}\cancel{(y - 9)}}{ \cancel{(y - 9)}\cancel{(y - 6)}(y + 2)} $ We are dividing by $y - 6$ , so $y - 6 \neq 0$ Therefore, $y \neq 6$ $k = \dfrac{90y}{y + 2} ; \space y \neq 9 ; \space y \neq 6 $